Can someone give me the answers

A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 181 ms. How

lara [203]

Answer:

the center of mass is 316,670m bellow the release point.

Explanation:

First, we must find the distance at which the objects are in time t = 360s

We will use the formula for vertical distance in free fall

$h=v_{0}t+\frac{1}{2} gt^2$

$v_{0}$ is the initial velocity, is 0 for both stones since they were just dropped.

g is acceleration of gravity and t is time ( $g=9.81m/s^2$ )

At $t=360s$ the first stone has been falling for the entire 360 seconds, its position h1 is:

$h_{1}=\frac{1}{2} (9.82m/s^2)(360s^2)=635,688m$

And at 360 seconds the second stone has been fallin for $t= 360s -181 s = 179s$ , so its position h2 is:

$h_{2}=\frac{1}{2} (9.81m/s^2)(179)^2=157,161.1m$

And finally using the equation for the center of mass:

$CM=\frac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}$

We know that the mass of the second stone is twice the mass of the first stone so:

$m_{1}=m\\m_{2}=2m$

replacing these values in the equation for the center of mass

$CM=\frac{mh_{1}+2mh_{2}}{m+2m}$

$CM=\frac{m(h_{1}+2h_{2})}{3m}=\frac{h_{1}+2h_{2}}{3}$

Finally, replacing the values we found fot h1 and h2:

$CM=\frac{635,688m+2(157,161.1m)}{3}=316,670m$

the center of mass is 316,670m bellow the release point.

8 0

4 years ago

Give two examples of direct current electricity.

ludmilkaskok [199]

Answer:

<h3>Ion beams and Electrochemical are two examples of direct current electricity.</h3>

6 0

4 years ago

A square current loop 4.9 cm on each side carries a 520 mA current. The loop is in a 1.0 T uniform magnetic field. The axis of t

dybincka [34]

Answer:

Torque = 0.25Nm

Explanation:

To calculate the magnitude of torque, first we must calculate the lever arm. The lever arm is perpendicular distance from the axis of rotation to the line of the action of the force.

Lever arm = r sin theta

Where r = 4.9cm= 0.049m

Theta = 30°

Lever arm = 0.049 sin30° = 0.245m

Magnitude of torque = rfsintheta

f = 1.0T

Magnitude of torque= 0.245×1.0 = 0.245Nm

To 2 significant figure=0.25Nm

4 0

3 years ago

Read 2 more answers

You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone

umka21 [38]

Answer:

0.37 m

Explanation:

Given :

Window height, $h_1$ = 1.27 m

The flowerpot falls 0.84 m off the window height, i.e.

$h_2$ = (1.27 x 0.84 ) m in a time span of $t=\frac{8}{30}$ seconds.

Assuming that the speed of the pot just above the window is v then,

$h_2=ut+\frac{1}{2}gt^2$

$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$

$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$

$v= 2.69$ m/s

Initially the pot was dropped from rest. So, u = 0.

If it has fallen from a height of h above the window then,

$h = \frac{v^2}{2g}$

$h = \frac{(2.69)^2}{2 \times 9.81}$

h = 0.37 m

3 0

3 years ago

Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds th

Oksi-84 [34.3K]

Answer:

The frequencies are $f_n = n (0.875 )$

Explanation:

From the question we are told that

The speed of the wave is $v = 0.700 \ m/s$

The length of vibrating clothesline is $L = 40.0 \ cm = 0.4 \ m$

Generally the fundamental frequency is mathematically represented as

$f = \frac{v}{2 L }$

=> $f = \frac{ 0.700 }{2 * 0.4 }$

=> $f = 0.875 \ Hz$

Now this other frequencies of vibration experience by the clotheslines are know as harmonics and they are obtained by integer multiple of the fundamental frequency

So

The frequencies are mathematically represented as

$f_n = n * f$

=> $f_n = n (0.875 )$

Where n = 1, 2, 3 ....

3 0

3 years ago